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If you note $\|\cdot \|_\infty$ the sup ess norm, you have : $$\| u - \tilde{u} \|_{\infty} \leq \| u - u_k\|_{\infty}+ \| u_k - \tilde{u} \|_{\infty}$$ Now let $\epsilon > 0$, As $u_k \to \tilde{u}$ uniformly, there exist $k_1$ such that $\forall k > k_1, \ \| \tilde{u} - u_k\|_{\infty} < \frac{\epsilon}{2}$ As $u_k \to u$ in $W^{1,2}_0$, it ...

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$$\frac {\partial K(x,y)}{\partial x}y - \frac {\partial K(x,y)}{\partial y}x =0$$ $$\frac {\partial K(x,y)}{x\partial x} - \frac {\partial K(x,y)}{y\partial y} =0$$ Let $X=x^2$ ; $Y=y^2$ ; $K(x,y)=H(X,Y)$ $$\frac {\partial H(X,Y)}{\partial X} - \frac {\partial H(X,Y)}{\partial Y} =0$$ The general solution of this PDE is well-known : $$H(X,Y)=H(X+Y)$$ any ...

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Matrix has icomplete rank, so the system is solvable only if $$\operatorname{rank} M = \operatorname{rank} \begin{pmatrix}M \;\big|\; f\end{pmatrix}.$$ Due to truncation error, the last may not hold (but would hold, if you're using a conservative approximation, I suppose). You can use QR decomposition to deal with that. Suppose $M = QR$ where $Q$ is ...

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A parabolic operator with constant coefficients is a linear transformation away from the heat operator, so it is a natural guess that the fundamental solutions should be similar. I will use this idea to find the fundamental solution. (If you just want to see the solution, see the last line.) Take two positive definite symmetric $n\times n$ matrices $A$ and ...

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Take a look at Couple stress theory for solids

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I would convert that into a system of two equations. Let $v = U_t$ and $w = -U_x$. Then you have a following pair of equations $$v_t + (1 + \epsilon w^2) w_x = 0\\ w_t + v_x = 0$$ These equations can be rewritten in conservative form as $$\frac{\partial \mathbf Z}{\partial t} + \frac{\partial \mathbf F(\mathbf Z)}{\partial x} = 0$$ with $\mathbf Z = (v, ... 1 Following the hint by Ian, let$v=u+1$. The function$v$satisfies the PDE$\Delta v=v$, and is positive on the boundary of$\Omega$. So, if$v\$ was negative somewhere, its (negative) global minimum would be attained in the interior... but the Laplacian can't be negative at an interior minimum.

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This is a classical result. You could find the answer on this book,page 176, section 5.2.3, theorem 4

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